Example of the method of characteristics

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Homework Statement


Hi,guys I have a example, i understand almost everything but i have problems understanding some steps.
Example:
2u't+3u'x=0,x\inR,t>0,u(x,0)=sin(x),u=u(x,y)


The Attempt at a Solution


I rewrite the example
(1)2u't+3u'x=u's
(2)u'tt's+u'xx's=u's

From comparing (1) (2) we can obtain
(3)t's=2
(4)x's=3

x=3s+C
t=2s+A
I suppose these equations are obtained by just integrating (3) and (4)

And here is the confusing par for me:
x,when s=0, =\eta
t,when s=0, =0
I do not understand how these conditions are made.

Using these conditions we can calculate constants C,A.
Then express new variables \eta and s.

The problem with respect to new variables:
u,when s=0, =sin(\eta)
u's=0

How I sad, i do not understand how conditions (3) and (4) are made and then the "new problem":
u,when s=0, =sin(\eta) is confusing ,it seems that s=t.
Could someone ,please,can explain this to me?
It would help a lot.

Homework Statement

 
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prehisto said:

Homework Statement


Hi,guys I have a example, i understand almost everything but i have problems understanding some steps.
Example:
2u't+3u'x=0,x\inR,t>0,u(x,0)=sin(x),u=u(x,y)

You have 3 variables ##t,x,y## on this line. And strange notation like ##u'_x##. So I am going to rewrite it in a form that makes sense to me:$$
2u_x(x,y) + 3u_y(x,y) = 0,~u(x,0) = \sin x$$

The Attempt at a Solution


I rewrite the example
(1)2u't+3u'x=u's
(2)u'tt's+u'xx's=u's

I guess, from what you wrote, that you are looking for a curve ##\vec R(s) = \langle x(s),y(s)\rangle## such that ##u(s) = u(x(s),y(s))## is a solution. So you want$$
u'(s) = \nabla u \cdot \frac{d\vec R}{ds}=u_x x'(s) + u_y y'(s) = 3u_x + 2u_y = 0$$

From here I will veer away from what you did. With ##x'(s)=3, y'(s)=2## you have the slope of the curve is ##\frac {dy}{dx}= \frac{y'(s)}{x'(s)}=\frac 2 3##. This says ##u(s)## is constant on lines with slope ##\frac 2 3##. So the value of ##u(x,y)## at any point in the plane is the value of ##u## where the line through ##(x,y)## with slope ##\frac 2 3## intercepts the ##x## axis. See if you can show that point would be ##(x-\frac 3 2 y, 0)##. And the value of ##u(x,y)## at points on the ##x## axis is given.

Hopefully this may be of some help to you. Disclaimer: I am not a PDE guy.
 
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